03. The Born Interpretation

3.1 Probability Density

Probability Density Function

Let $\Psi (\mathbf{x},t)$ be particle’s wave function

Probability Density Function for position of particle is

$$\rho (\mathbf{x},t)\equiv |\Psi (\mathbf{x},t){|}^2$$

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Probability of Particle in Region

Let $\Psi (\mathbf{x},t)$ be the particle’s wave function

Probability of finding particle in volume $D\subset {\mathbb{R}}^3$ is

$${\mathbb{P}}_{\Psi }(D)={\iiint }_D|\Psi (\mathbf{x},t){|}^2{d}^{3}x$$

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Normalisable Wave Function

Wave Function $\Psi$ is normalisable if

$$0<{\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x<\infty ,\forall \text{ time }t$$

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Normalised Wave Function

Wave function $\Psi$ is normalised if

$${\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x=1,\forall \text{ time }t$$

One-Dimensional Version

$${\int }_{-\infty }^{\infty }|\Psi (x,t){|}^{2}dx=1$$

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Normalised Particle in Box

$${\Psi }_n(x,t)=\{\begin{array}{ll}\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)e^{-i{n}^2{\pi }^2\hslash t/2m{a}^2}& 0<x<a\\ 0& \text{otherwise}\end{array}$$

Proof

$${\int }_{-\infty }^{\infty }|{\Psi }_n(x,t){|}^{2}dx=|B{|}^2{\int }_0^a{\sin }^2\left(\frac{n\pi x}{a}\right)dx=\frac{1}{2}a|B{|}^2$$

So $B=\sqrt{\frac{2}{a}}$ normalises

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Normalisation for Stationary State Wave Functions of energy $E$

Let $\Psi$ be a Stationary state wave function of energy $E$

Then

$$\rho (\mathbf{x},t)=|\Psi (\mathbf{x},t){|}^2=|\psi (\mathbf{x})e^{-iEt/\hslash }{|}^2=|\psi (\mathbf{x}){|}^2$$

Hence $\Psi (\mathbf{x},t)$ is normalised for all time if $\varphi (\mathbf{x})$ is normalised for all time so requires

$${\iiint }_{{\mathbb{R}}^3}|\psi (\mathbf{x}){|}^2{d}^{3}x=1$$

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Distribution Function

Let $\rho (\mathbf{x})$ be the Probability Density Function then

Distribution Function is defined as

$$F(x)\equiv {\int }_{-\infty }^{\infty }\rho (y)dy$$

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Probability Density and Distribution Function for Particle in a Box

Let $\Psi$ be wave function for Particle in a box

$${\rho }_n(x)=\frac{2}{a}{\sin }^2\left(\frac{n\pi x}{a}\right)=\frac{1}{a}\left(1-\cos \left(\frac{2n\pi x}{a}\right)\right)$$ $$F_n(x)\equiv {\int }_0^x{\rho }_n(y)dy=\frac{x}{a}-\frac{1}{2n\pi }\sin \left(\frac{2n\pi x}{a}\right)$$

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Correspondence Principle

Tendency of Quantum Results to approach Classical Theory for Large Quantum Numbers

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Expectation Value of Function of Position

Let $f(\mathbf{x})$ be a function of position
Let $\Psi (\mathbf{x},t)$ be wave function satisfying Schrödinger Equation

Expectation Value of Function of Position $f(\mathbf{x})$ is

$${\mathbb{E}}_{\Psi }(f(\mathbf{x}))={\iiint }_{{\mathbb{R}}^3}f(\mathbf{x})|\Psi (\mathbf{x},t){|}^2{d}^{3}x$$

One Dimension Version

$${\mathbb{E}}_{\Psi }(f(x))\equiv {\iiint }_{\infty }^{\infty }f(x)|\Psi (x,t){|}^{2}dx$$

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Expectation of a Position for Particle in a Box

Let $\Psi$ be wave function satisfying Particle in a Box

$$\begin{array}{rl}{\mathbb{E}}_{{\Psi }_n}(x)& ={\int }_0^{a}x{\rho }_n(x)dx\\ & =[x{F}_n(x){]}_0^a-{\int }_0^a{F}_n(x)dx\\ & =a-{\left[\frac{x^2}{2a}+\frac{a}{4n^2{\pi }^2}\cos \left(\frac{2n\pi x}{a}\right)\right]}_0^a\\ & =\frac{1}{2}a\end{array}$$

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3.2 The Continuity Equation and Boundary Conditions

Continuity Equation for Schrödinger Equation

Schrödinger Equation implies continuity equation

$$\frac{\partial \rho }{\partial t}+\nabla \cdot \mathbf{j}=0$$

where $\rho (\mathbf{x},t)=|\Psi (\mathbf{x},t){|}^2$ and vector field

$$\mathbf{j}(\mathbf{x},t)\equiv \frac{i\hslash }{2m}(\Psi (\mathbf{x},t)\cdot \overline{(\nabla \Psi )(\mathbf{x},t)}-\overline{\Psi (\mathbf{x},t)}\cdot (\nabla \Psi )(\mathbf{x},t))$$

is known as the probability current $\mathbf{j}$

Proof

$$\begin{array}{rl}\frac{\partial }{\partial t}|\Psi (\mathbf{x},t){|}^2& =\left(\frac{\partial }{\partial t}\overline{\Psi (\mathbf{x},t)}\right)\Psi (\mathbf{x},t)+\overline{\Psi (\mathbf{x},t)}\frac{\partial }{\partial t}\Psi (\mathbf{x},t)\\ & =\overline{\left[-\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\right]}\Psi +\overline{\Psi }\left[-\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\right]\\ & =\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\overline{\Psi }+V\overline{\Psi }\right)\Psi -\overline{\Psi }\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\\ & =\frac{i\hslash }{2m}(\overline{\Psi }{\nabla }^2\Psi -\Psi {\nabla }^2\overline{\Psi })\\ & =\frac{i\hslash }{2m}\nabla \cdot (\overline{\Psi }\nabla \Psi -\Psi \nabla \overline{\Psi })=-\nabla \cdot \mathbf{j}\end{array}$$

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Conservation of Probability

Suppose for all time $t$,

$\mathbf{j}(\mathbf{x},t)$ satisfies boundary condition that it tends to zero faster than $\frac{1}{|\mathbf{x}{|}^2}$ as $|\mathbf{x}|=r\to \infty$

Hence

$${\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x\text{ is independent of }t$$

In particular, if $\Psi$ is normalised at some time $t=t_0$ then it is normalised for all $t$

Proof

Let $S$ be a closed surface that bounds region $D\subset {\mathbb{R}}^3$ then

$$\begin{array}{rl}\frac{d}{dt}{\iiint }_D|\Psi (\mathbf{x},t){|}^2{d}^{3}x& ={\iiint }_D\frac{\partial \rho }{\partial t}{d}^{3}x\\ & ={\iiint }_D(-\nabla \cdot \mathbf{j})d^{3}x\\ & =-{\iint }_S\mathbf{j}\cdot d\mathbf{S}\end{array}$$

Suppose $S$ is a sphere of radius $r>0$, centred on origin, so $D$ is a ball
Then outward pointing unit normal vector to $S$ is

$$\mathbf{n}=\frac{\mathbf{x}}{r}$$

Hence

$$\mathbf{j}\cdot d\mathbf{S}=\mathbf{j}\cdot \mathbf{n}{r}^{2}d{S}_{\text{unit}}$$

where
$d{S}_{\text{unit}}=\sin \theta d\delta d\varphi$ is area element on unit radius sphere in spherical polar coordinates

As $\mathbf{j}\cdot \mathbf{n}=o\left(\frac{1}{r^2}\right)$ uniformly in angular coordinates then

Hence as $r\to \infty$ then surface integral tends to zero hence

$$\frac{d}{dt}{\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x=0$$

Thus

$${\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x\text{ is independent of }t$$

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Conditions for Solutions to Schrödinger Equation

1. Wave Function $\Psi (\mathbf{x},t)$ should be continuous, single-valued function
   Ensures probability density function $\rho =|\Psi {|}^2$ is single-valued with no discontinuities

2. $\Psi$ should be normalisable
   Ensures integral of $|\Psi {|}^2$ over all space should be finite and non-zero
   Note that this condition may be relaxed for free particles

3. $\nabla \Psi$ should be continuous everywhere, except at an infinite discontinuity in potential $V$

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3.3 Measurement of Energy

Initial Value Problem for Schrodinger's Equation for Particle in Box

Consider $\Psi (x,0)=f(x)$ where $f:[0,a]\to \mathbb{C}$ with $f(0)=f(a)=0$

$$\Psi (\mathbf{x},t)=\sum _{n=1}^{\infty }{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$$

where

$$c_m={\int }_0^{a}f(x){\psi }_m(x)dx$$

Proof $f(0)=f(a)=0$ then $f$ can be expanded as a Fourier Sine Series

As

$$f(x)=\sum _{n=1}^{\infty }{c}_n\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)=\sum _{n=1}^{\infty }{c}_n{\psi }_n(x)$$

for appropriate $c_n\in \mathbb{C}$

As ${\psi }_n(x)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)$ are orthonormal

$${\int }_0^a{\psi }_m(x){\psi }_n(x)dx={\delta }_{mn}$$

Hence

$${\int }_0^{a}f(x){\psi }_m(x)dx=\sum _{n=1}^{\infty }{c}_n{\int }_0^a{\psi }_n(x){\psi }_m(x)dx=c_m$$

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Probability of Measuring Energy of Particle

Suppose IVP for Schrodinger’s Equation for particle in box is normalised then

Probability of Measuring energy of particle to be $E_n$ is $|c_n{|}^2$

Proof

$$1={\int }_0^a|\Psi (x,t){|}^{2}dx=\sum _{m,n=1}^{\infty }\overline{c_m}{c}_n{e}^{-i(E_n-E_m)t/\hslash }{\int }_0^a\overline{{\psi }_m}(x){\psi }_n(x)dx=\sum _{n=1}^{\infty }|c_n{|}^2$$

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